Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& i : v) is >> i;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
  for (auto& i : v) os << i << ' ';
  return os;
}
template <typename T, typename U>
istream& operator>>(istream& is, pair<T, U>& p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, pair<T, U>& p) {
  os << p.first << ' ' << p.second;
  return os;
}
template <typename T, typename U>
pair<T, U> operator-(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
template <typename T, typename U>
pair<T, U> operator+(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
const int nax = 100005;
const int naxl = 20;
int n, m, k;
vector<vector<int>> g;
vector<bool> dead;
array<int, nax> sz;
array<set<int>, nax> cnt;
array<array<pair<int, set<int>>, nax>, naxl> cent;
int dfs_sz(int v, int p = -1) {
  sz[v] = 1;
  for (auto u : g[v])
    if (u != p && !dead[u]) sz[v] += dfs_sz(u, v);
  return sz[v];
}
int find_centroid(int v, int s, int p = -1) {
  for (auto u : g[v])
    if (!dead[u] && u != p && sz[u] * 2 > s) return find_centroid(u, s, v);
  return v;
}
void dfs_set(int v, int c, int lvl, set<int> s, int p = -1) {
  for (auto y : cnt[v]) s.insert(y);
  while (s.size() > 10) s.erase(--s.end());
  cent[lvl][v] = {c, s};
  for (auto u : g[v])
    if (!dead[u] && u != p) dfs_set(u, c, lvl, s, v);
}
void build(int v = 0, int lvl = 0) {
  int c = find_centroid(v, dfs_sz(v));
  dead[c] = true;
  dfs_set(c, c, lvl, set<int>());
  for (auto u : g[c])
    if (!dead[u]) build(u, lvl + 1);
}
set<int> ask(int u, int v) {
  int lvl = 0;
  while (cent[lvl][u].first == cent[lvl][v].first) ++lvl;
  --lvl;
  set<int> s = cent[lvl][u].second;
  for (auto y : cent[lvl][v].second) s.insert(y);
  return s;
}
int main() {
  cin >> n >> m >> k;
  g.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < m; ++i) {
    int u;
    cin >> u;
    cnt[u - 1].insert(i + 1);
  }
  dead.assign(n, false);
  for (int i = 0; i < naxl; ++i) {
    cent[i].fill(make_pair(-1, set<int>()));
  }
  build();
  42;
  ;
  for (int i = 0; i < k; ++i) {
    int v, u, a;
    cin >> v >> u >> a;
    if (u == v) {
      if (!cnt[u - 1].empty()) {
        cout << min(a, (int)cnt[u - 1].size());
        auto it = cnt[u - 1].begin();
        for (int i = 0; i < min(a, (int)cnt[u - 1].size()); ++i) {
          cout << ' ' << *it;
          ++it;
        }
        cout << '\n';
      } else
        cout << 0 << '\n';
      continue;
    }
    set<int> s = ask(v - 1, u - 1);
    while (s.size() > a) s.erase(--s.end());
    cout << s.size();
    for (auto u : s) {
      cout << ' ' << u;
    }
    cout << '\n';
  }
  return 0;
}